Hi…wheneva calculations for any problem are difficult, the options are friendly…hence, if v sumhow manage to find an approximate value of a difficult calculation, we can trace the correct option… *_marking a question correctly does not …

Hi...wheneva calculations for any problem are difficult, the options are friendly...hence, if v sumhow manage to find an approximate value of a difficult calculation, we can trace the correct option...

** marking a question correctly does not necessarily require getting the correct answer**...

if v can leverage the gap between options...we can save few precious seconds..

lets learn this art of finding approx results

**Multiplication.**

suppose v need to do an actual brutal calculation...lemme jot down few huge nos. randomly...

5467398 x 345678 x 6578609 x 4354657x 345267 x 768907

v know this wont appear in cat...but v say in army (my earlier organisation)...if u're ready for the most difficult situation, u're obviously ready for the easier ones....

step 1..most important...

take the powers away...n forget wateva's there after first place of decimal...it'll have minimal effect on the final answer...

i.e. express the above figures as...

5.4 x 3.4 x 6.6 x 4.4 x 3.4 x 7.7 x 10^ 33

why is it important?

this step wud reduce the error...u've taken bulk of the quantity away...so wateva error wud be there wud be in multiplication of small quantities like 5.4 n 3.4...

step 2..

round of values...try rounding off one value on lower side...other on the higher side for better approximation...

e.g. 99 x 11...increase 99 by 1 and reduce 11 by 1 so 99.x 11 becomes 100x10 = 1000.

the accurate value of 9.9 x 1.1 is 1089...not much difference...but time saved is huge...

if u can remember that by this method, the value obtained is a bit less (on avg 5%), than the actual value...ur approximation wud be better...

lets get back to our question...

we write the expression as 5x4x7x4x3x8 = 140 x 96 = aprrox 140 x 100 = 14000

deliberately took 96 as 100 as v know answer obtained is a bit smaller...it compensated 4% of the average -5% error...the actual value wud be slightly more...say 14200..

.

so, including power, value is approx. 1.42 x 10^37

well...the correct answer is 1.438 x 10^ 37

lets take one more example....this time sumthing that may appear in cat...

say...323 x 87.

looks a cake walk? it is!

3.2 x 8.7 x 10^3...

3 x 9

= 27000

we know answer wud be a bit more than this....approx. 28000..

correct answer....28101.

i'll recommend u to write few crude values right now and practice this a bit...

**Division**

again...lets take sum brutal value...3452673/746372276354...

step 1 & 2 simultaneously...same as above..take the power out...and round off the first 2 digits of each quantity...

35 x 10^5 / 7.5 x 10^11

it becomes 35/7.5 x 10^-6 (took 35 n not 3.5 to make it bigger than denominator..its always easy to divide a bigger no. by smaller no.)

step 3...read it carefully...

this time v will either decrease both figures or increase both figures...bcoz in division...nos. are in opposition...v either decrease both or increase both to put up a similar fight...(in multiplication, nos. are supporting each other...so if v increase one, v shud decrease other to maintain same strength)

another thing...35 is 4-5 times 7.5...so wateva value v add or subtract to 3.5, v'll add/subtract 4-5 times that value to 35 for better calculatn.

v made 7.5 => 8...increments of .5

so v'll increase 35 by 2...

so the value is 37/8 = 4+ 5/8 = 4.625

including power the answer becomes...4.625 x 10^-6

correct answer is...

will u believe...4.622 x 10^-6.

lets now take a cat level example....

say.... 345/43526.

34/4.3 x 10^-3

4.3 becomes 4....decrement of .3 since 34 is approx 7 times...v decrease 3x.7 = 2.1 to 34...it becomes 32..

so the value is 32/ 4 x 10^ -3

= 8 x 10^-3

with more practice...u'll become actually quick...for example...i cud see 4 is a bit more than 4 and 34 is a bit more than 32 (which is 4x8 ..so the value will be more or less 8 only.

or, 4x8 = 32, .3x8 = 2.4 sum = 34.4...no. with us is 34...so the value shud be a bit less than 8...what say?

the answer is 7.93 x 10^...

**percentage**

if u have mastered X and /...% is a child's play...its application of both together.

important:

always think of percentage as decimals... 10% is like .1 of the value...

growth of 10 % = 1.1 times the value

growth of 30 % = 1.3 times the value

decrement of 30 % = .7 times the value

decrement to 40% = .4 times the value

% change = (new value-old value)/old value x 100

lets take sum calculations....say 35 % of 23.7

v need to find out .35 x 23.7 it becomes a simple X problem...right?

if the answer is in %, assume total qtty to be 100.

e.g. price has grown by 25%, by what % shud consumptn be reduced to avoid a hike in expenditure?

exp. = price x consumptn. change in consumptn is to be determined....take it as 100.

and let the price be 1/- coz 1 is easiest no. to multiply (after 0...:-P )

so earlier expenditure was 1 x 100 = 100

now also expenditure shud be 100... but price has become 1x1.25=1.25. so the consumptn shud become 100/1.25 (new exp./new price) = 80

100 has become 80...so a decrement of 20%.

**square root:**

a sincere request...please done eva, eva calculate square root by conventional method...look at the method below...with lil sincerity...it will yield answers with less than 1 % accuracy...lets have a look...

sqrt(6567890876323577654)

= sqrt(6.6 x 10^18 ...please not that i took even power of 10...so tht i can take it out.

= 10^9 x sqrt(6.6)

now, 2^2 = 4

3^2 = 9

6.6 lies between 4 n 9 so the answer shud be...

2.#$ x 10 ^ 9

in most of the cases, u'd know the answer by this much...still...if u want to get places after decimal...watch it carefully...

difference between two squares = 9-4 = 5 (denominator)

difference between no. v r targeting and the smaller square... = 6.6 - 4 = 2.4 (numerator)

the value after decimal will be 2.4/5 = .48

so the value of sqrt is 2.48 x 10^9

again...remember while this process (interpolation) we have assumed a parabola to behave like a line...which it does between two very close value....the answer wud be a bit-bit greater than what v've calculated...

e.g. if we've calculated 2.48, it wud be around 2.53 x 10^9. (u can always add values b/w .02 n .08 to get actual result...even if u dont do this, the answer remains pretty close)

in between....the correct ans is 2.56 x...

lets take a cat level example...

say..sqrt(8582)

10 x sqrt(85.

9^2 = 81

10^2 = 100

ans is

**9**.&$#@...if its insufficient...

value after decimal = (85.8-81)/(100-81) = 5/20 = .25

value becomes 9.25...include a correction of .05...it becomes 9.3 x 10 = 93

ans is 92.64

few things to be kept in mind...

#1. the level of accuracy of ur calculation shud depend on gaps b/w options...if options are very far off...go for very casual approximation...else...ponder a bit.

#2. not every method suits everybody...so please dont bother too much if sum method doesnt suit u.

#3. the methods u like shud become a part of ur day to day calculations....

For me....calculations is a hobby...right from my job interview to my IIM interview to my army interview...i have faced questions on same...the methods described here are mine...its not vedic...i jus say its love with numbers...thats y i had to open a new thread...in case u guys find it interesting/helpful...i can share lotsa other methods for various other calculations as well...n wud get an opportunity to learn from the great brains around....

all the best puys...

with warm regards,

maxximus

thanx.....its really a usefull mattar...........

one really need to save time using these fundas

thanxquote=maxximus;781298]Hi...wheneva calculations for any problem are difficult, the options are friendly...hence, if v sumhow manage to find an approximate value of a difficult calculation, we can trace the correct option...

** marking a question correctly does not necessarily require getting the correct answer**...

if v can leverage the gap between options...we can save few precious seconds..

lets learn this art of finding approx results

**Multiplication.**

suppose v need to do an actual brutal calculation...lemme jot down few huge nos. randomly...

5467398 x 345678 x 6578609 x 4354657x 345267 x 768907

v know this wont appear in cat...but v say in army (my earlier organisation)...if u're ready for the most difficult situation, u're obviously ready for the easier ones....

step 1..most important...

take the powers away...n forget wateva's there after first place of decimal...it'll have minimal effect on the final answer...

i.e. express the above figures as...

5.4 x 3.4 x 6.6 x 4.4 x 3.4 x 7.7 x 10^ 33

why is it important?

this step wud reduce the error...u've taken bulk of the quantity away...so wateva error wud be there wud be in multiplication of small quantities like 5.4 n 3.4...

step 2..

round of values...try rounding off one value on lower side...other on the higher side for better approximation...

e.g. 99 x 11...increase 99 by 1 and reduce 11 by 1 so 99.x 11 becomes 100x10 = 1000.

the accurate value of 9.9 x 1.1 is 1089...not much difference...but time saved is huge...

if u can remember that by this method, the value obtained is a bit less (on avg 5%), than the actual value...ur approximation wud be better...

lets get back to our question...

we write the expression as 5x4x7x4x3x8 = 120 x 12 x 96 = 1440 x 100 = 144000

deliberately took 96 as 100 as v know answer obtained is a bit smaller...it compensated those 5%.

so, including power, value is approx. 1.44 x 10^38

well...the correct answer is 1.438 x 10^ 38.

lets take one more example....this time sumthing that may appear in cat...

say...323 x 87.

looks a cake walk? it is!

3.2 x 8.7 x 10^3...

3 x 9

= 27000

we know answer wud be a bit more than this....approx. 28000..

correct answer....28101.

i'll recommend u to write few crude values right now and practice this a bit...

**Division**

again...lets take sum brutal value...3452673/746372276354...

step 1 & 2 simultaneously...same as above..take the power out...and round off the first 2 digits of each quantity...

35 x 10^5 / 7.5 x 10^11

it becomes 35/7.5 x 10^-6 (took 35 n not 3.5 to make it bigger than denominator..its always easy to divide a bigger no. by smaller no.)

step 3...read it carefully...

this time v will either decrease both figures or increase both figures...bcoz in division...nos. are in opposition...v either decrease both or increase both to put up a similar fight...(in multiplication, nos. are supporting each other...so if v increase one, v shud decrease other to maintain same strength)

another thing...35 is 4-5 times 7.5...so wateva value v add or subtract to 3.5, v'll add/subtract 4-5 times that value to 35 for better calculatn.

v made 7.5 => 8...increments of .5

so v'll increase 35 by 2...

so the value is 37/8 = 4+ 5/8 = 4.625

including power the answer becomes...4.625 x 10^-6

correct answer is...

will u believe...4.622 x 10^-6.

lets now take a cat level example....

say.... 345/43526.

34/4.3 x 10^-3

4.3 becomes 4....decrement of .3 since 34 is approx 7 times...v decrease 3x.7 = 2.1 to 34...it becomes 32..

so the value is 32/ 4 x 10^ -3

= 8 x 10^-3

with more practice...u'll become actually quick...for example...i cud see 4 is a bit more than 4 and 34 is a bit more than 32 (which is 4x8 ..so the value will be more or less 8 only.

or, 4x8 = 32, .3x8 = 2.4 sum = 34.4...no. with us is 34...so the value shud be a bit less than 8...what say?

the answer is 7.93 x 10^...

**percentage**

if u have mastered X and /...% is a child's play...its application of both together.

important:

always think of percentage as decimals... 10% is like .1 of the value...

growth of 10 % = 1.1 times the value

growth of 30 % = 1.3 times the value

decrement of 30 % = .7 times the value

decrement to 40% = .4 times the value

% change = (new value-old value)/old value x 100

lets take sum calculations....say 35 % of 23.7

v need to find out .35 x 23.7 it becomes a simple X problem...right?

if the answer is in %, assume total qtty to be 100.

e.g. price has grown by 25%, by what % shud consumptn be reduced to avoid a hike in expenditure?

exp. = price x consumptn. change in consumptn is to be determined....take it as 100.

and let the price be 1/- coz 1 is easiest no. to multiply (after 0...:-P )

so earlier expenditure was 1 x 100 = 100

now also expenditure shud be 100... but price has become 1x1.25=1.25. so the consumptn shud become 100/1.25 (new exp./new price) = 80

100 has become 80...so a decrement of 20%.

**square root:**

a sincere request...please done eva, eva calculate square root by conventional method...look at the method below...with lil sincerity...it will yield answers with less than 1 % accuracy...lets have a look...

sqrt(6567890876323577654)

= sqrt(6.6 x 10^18 ...please not that i took even power of 10...so tht i can take it out.

= 10^9 x sqrt(6.6)

now, 2^2 = 4

3^2 = 9

6.6 lies between 4 n 9 so the answer shud be...

2.#$ x 10 ^ 9

in most of the cases, u'd know the answer by this much...still...if u want to get places after decimal...watch it carefully...

difference between two squares = 9-4 = 5 (denominator)

difference between no. v r targeting and the smaller square... = 6.6 - 4 = 2.4 (numerator)

the value after decimal will be 2.4/5 = .48

so the value of sqrt is 2.48 x 10^9

again...remember while this process (interpolation) we have assumed a parabola to behave like a line...which it does between two very close value....the answer wud be a bit-bit greater than what v've calculated...

e.g. if we've calculated 2.48, it wud be around 2.53 x 10^9. (u can always add values b/w .02 n .08 to get actual result...even if u dont do this, the answer remains pretty close)

in between....the correct ans is 2.56 x...

lets take a cat level example...

say..sqrt(8582)

10 x sqrt(85.

9^2 = 81

10^2 = 100

ans is

**9**.&$#@...if its insufficient...

value after decimal = (85.8-81)/(100-81) = 5/20 = .25

value becomes 9.25...include a correction of .05...it becomes 9.3 x 10 = 93

ans is 92.64

few things to be kept in mind...

#1. the level of accuracy of ur calculation shud depend on gaps b/w options...if options are very far off...go for very casual approximation...else...ponder a bit.

#2. not every method suits everybody...so please dont bother too much if sum method doesnt suit u.

#3. the methods u like shud become a part of ur day to day calculations....

For me....calculations is a hobby...right from my job interview to my IIM interview to my army interview...i have faced questions on same...the methods described here are mine...its not vedic...i jus say its love with numbers...thats y i had to open a new thread...in case u guys find it interesting/helpful...i can share lotsa other methods for various other calculations as well...n wud get an opportunity to learn from the great brains around....

all the best puys...

with warm regards,

maxximus

Excellent initiative maxximus .

This is one very important aspect which we need to take care of , for competitive exams .

Keep pouring in 😃

we write the expression as 5x4x7x4x3x8 = 120 x 12 x 96 = 1440 x 100 = 144000

all the best puys...

with warm regards,

maxximus

hi maxximus,

how do we get 120*12*96 from 5*4*7*4*3*8?

thanks

Great stuff maxximus ... I wonder why so many books on CAT are shy of these simple but effective techniques ?!

great !!! brilliant work.

maxx , extremely good initiative bro....continue with new methods...these big calculations really consumes a lot of time...

great going dude

NIce thread